$\overline{AB}$ = $\sqrt{149}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $\sqrt{149}$ $?$ $ \sin( \angle ABC ) = \frac{7\sqrt{149} }{149}, \cos( \angle ABC ) = \frac{10\sqrt{149} }{149}, \tan( \angle ABC ) = \dfrac{7}{10}$
Answer: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is opposite to $\angle ABC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle ABC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{149}} $ $ \overline{AC}=\sqrt{149} \cdot \sin( \angle ABC ) = \sqrt{149} \cdot \frac{7\sqrt{149} }{149} = 7$